Does the bounded extension of the Fourier multiplier operator agrees with its original explicit definition? We consider the Fourier multiplier operator $T_0$ defined by the explicit expression
$$(T_0f)(x)=\int_{\mathbb{R}^n}{e^{ix\cdot \xi}m(\xi)\hat{f}(\xi)d\xi}, \ f\in S(\mathbb{R}^n),$$ where $S(\mathbb{R}^n)$ is the Schwartz function space. Here we assume that the multiplier $m(\xi)\in L^\infty(\mathbb{R}^n)$ satisfies the conditions in the H\"ormander's multiplier theorem, which implies that $T_0$ can be extended to a bounded operator $T$ from $L^p(\mathbb{R}^n)$ to $L^p(\mathbb{R}^n)$, $1<p<\infty$. Then it is natural to ask the following question. Do we have 
\begin{equation}(1)\quad\quad\quad (Tf)(x)=\int_{\mathbb{R}^n}{e^{ix\cdot \xi}m(\xi)\hat{f}(\xi)d\xi},\ a.e., \ f\in L^p(\mathbb{R}^n)\cap L^1(\mathbb{R}^n),\end{equation} whenever $|m(\xi)\hat{f}(\xi)|\in L^1(\mathbb{R}^n)$?
 A: We construct a sequence $f_k\in S(\mathbb{R}^n)$ s.t. $||f_k-f||_p\to 0$ and $$T_0f_k-\int_{\mathbb{R}^n}e^{ix\cdot\xi}m(\xi)\hat{f}(\xi)d\xi\to 0,\ a.e..$$ By the bounded extension of the multiplier operator, there is a function $g\in L^p$ s.t. $||T_0f_k- g||_p\to 0$. Then there is a subsequence $T_0f_{k_j}-g\to 0,\ a.e.$. Hence, $g=\int_{\mathbb{R}^n}e^{ix\cdot\xi}m(\xi)\hat{f}(\xi)d\xi$, which gives the desired equality.
The construction of the sequence: Let $\phi\in C_0^\infty$ and $\varphi=\hat{\phi}$. $\phi(0)=1$. $\varphi_\epsilon(x)=\epsilon^{-n}\varphi(x/\epsilon)$, $\phi_\delta=\phi(\delta x)$. Let $f_{\epsilon,\delta}=(\varphi_\epsilon\ast f)\phi_\delta$. Since
$$||f_{\epsilon,\delta}-f||_p\le||(\varphi_\epsilon\ast f)\phi_\delta-\varphi_\epsilon\ast f||_p+||\varphi_\epsilon\ast f-f||_p, $$
we can choose some $\epsilon_k$ and $\delta_k$ to make $||f_{\epsilon_k,\delta_k}-f||_p$ smaller than $1/k$.
We also need to show that $$|\int e^{ix\cdot \xi}m(\xi)(\hat{f_{\epsilon,\delta}}-\hat{f})d\xi|$$ can be small. 
We see that $$\int|m||\hat{f_{\epsilon,\delta}}-\hat{f}|\le \int|m||(1-\hat{\varphi_\epsilon})\hat{f}|+\int|m||h_\epsilon-h_\epsilon\ast\hat{\phi_\delta}|=I_1+I_2,$$ where $h_\epsilon=\hat{\varphi_\epsilon}\hat{f}$. Since $$|m||(1-\hat{\varphi_\epsilon})\hat{f}|\le |m\hat{f}|\in L^1,$$by dominated convergence, we can choose a subsequence $\epsilon_{k_j}$ s.t. $I_1\le 1/k$.
Fix $\epsilon=\epsilon_{k_j}$. We claim that $$|h_{\epsilon}\ast\hat{\phi_\delta})(\xi)|=|\int h_\epsilon(\xi-\delta y)\hat{\phi}(y)dy|\le C_{\epsilon,N}(1+|\xi|)^{-N},$$ where $C_{\epsilon,N}$ is independent of $\delta$. To see this, we need to use the facts that $||h_\epsilon||_\infty\le ||f||_1$ and ${\rm supp}\ h_\epsilon\subset B(0,\epsilon^{-1})$. Indeed, when $|\xi|\le 10/\epsilon$, $$\int |h_\epsilon(\xi-\delta y)\hat{\phi}(y)|dy\le C\int|\hat{\phi}(y)|dy\le C,$$
and when $|\xi|\ge 10/\epsilon$, $$\int |h_\epsilon(\xi-\delta y)\hat{\phi}(y)|dy\le C\int_{|\xi-\delta y|\le \epsilon^{-1}}|\hat{\phi}(y)|dy\le C_N\int_{|y|\ge (|\xi|-\epsilon^{-1})/\delta}|y|^{-N}dy \le C_N|\xi|^{n-N}.$$
This proves the claim.
So $$|m||h_\epsilon-h_\epsilon\ast\hat{\phi_\delta}|\le |m\hat{f}|+C_{\epsilon,N}|m|(1+|\xi|)^{-N}\in L^1.$$
Since $h_\epsilon$ is bounded and continuous and $\epsilon=\epsilon_{k_j}$ is fixed, we have
 $${\rm limsup}_{\delta\to 0}|h_\epsilon-h_\epsilon\ast\phi_\delta|(\xi)\le {\rm limsup}_{\delta \to 0}\int |h_\epsilon(\xi-\delta y)-h_\epsilon(\xi)||\hat{\phi}(y)|dy=0,$$
then by dominated convergence, we can choose a subsequence $\delta_{k_{j_l}}$ s.t.$I_2\le 1/k$. Now, the sequence $f_{\epsilon_{k_{j}}, \delta_{k_{j_l}}}$ is what we need.
Any comments are welcome:)
